When Two Samples of Water are Mixed, what Final Temperature Results?

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Worksheet #2

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Example #1: Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C.

This is problem 8a from Worksheet #2.

First some discussion, then the solution. Forgive me if the points seem obvious:

1) The colder water will warm up (heat energy "flows" into it). The warmer water will cool down (heat energy "flows" out of it).
2) The whole mixture will wind up at the SAME temperature. This is very, very important.
3) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water)

This problem type becomes slightly harder if a phase change is involved. For this example, no phase change. What that means is that only the specific heat equation will be involved

Solution Key Number One: We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.

The warmer water goes down from to 46.8 to x, so this means its Δt equals 46.8 minus x. The colder water goes up in temperature, so its Δt equals x minus 14.9.

That last paragraph may be a bit confusing, so let's compare it to a number line:

To compute the absolute distance, it's the larger value minus the smaller value, so 46.8 to x is 46.8 minus x and the distance from x to 14.9 is x - 14.9.

These two distances on the number line represent our two Δt values:

a) the Δt of the warmer water is 46.8 minus x
b) the Δt of the cooler water is x minus 14.9

Solution Key Number Two: the energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

qlost = qgain

However:

q = (mass) (Δt) (Cp)

So:

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

With qlost on the left side and qgain on the right side.

Substituting values into the above, we then have:

(32.2) (46.8 - x)(4.184) = (32.2) (x - 14.9) (4.184)

Solve for x


Example #2: Determine the final temperature when 45.0 g of water at 20.0 °C mixes with 22.3 grams of water at 85.0 °C.

Solution:

We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.

The warmer water goes down from to 85.0 to x, so this means its Δt equals 85.0 minus x. The colder water goes up in temperature (from 20.0 to the ending temperature), so its Δt equals x minus 14.9.

That last paragraph may be a bit confusing, so let's compare it to a number line:

To compute the absolute distance, it's the larger value minus the smaller value, so 85.0 to x is 85.0 minus x and the distance from x (the larger value) to 20.0 (the smaller value) is x - 20.0.

The energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

qlost = qgain

So, by substitution, we then have:

(22.3) (85.0 - x)(4.184) = (45.0) (x - 20.0) (4.184)

Solve for x


Example #3: Determine the final temperature when 30.0 g of water at 8.00 °C mixes with 60.0 grams of water at 28.2 °C.

Solution:

(60.0) (28.2 - x)(4.184) = (30.0) (x - 8.00) (4.184)


Example #4: A 29.5 g sample of methanol at 208.9 K is mixed with 54.3 g of methanol at 302.3 K. Calculate the final temperature of the mixture assuming no heat is lost to the containers and surroundings. The specific heat of methanol is 2.53 J g¯11

Solution:

Let the final temperature be 'x.' Therefore, the Δt for the warmer methanol will be '302.3 - x' and for the colder methanol, it is 'x - 208.9.' Remember, 'x' is the final temperature and it is lower than the warmer methanol and higher than the colder methanol.

Remember:

(1) (mass) (Δt) (Cp) = (mass) (Δt) (Cp)

(2) qlost on the left; qgain on the right.

Substituting and solving, we have:

(29.5) (x - 208.9) (2.53) = (54.3) (302.3 - x) (2.53)

29.5x - 6162.55 = 16414.89 - 54.3x

83.8x = 22577.44

x = 269.4 K

In case, you're not sure what happened to the 2.53, I simply divided both sides by 2.53 first.


Example #5: A sheet of nickel weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

Solution:

This problem requires us to find the specific heats for nickel and iron. To do this, we will use this site. The values given are respectively, 0.54 J g¯1 °C¯1 and 0.46 J g¯1 °C¯1

Notice that the units on the site are kJ kg¯11. In addition, notice that I wrote J g¯1 °C¯1. Also, notice that there is no numerical difference when using either specific heat unit (the kJ one or the J one). In other words:

one kJ kg¯11 = one J g¯1 °C¯1

The left-hand unit is the IUPAC-approved one; the one on the right-hand side is the one in most common use.

On to the solution:

qlost = qgain

Therefore:

(20.0) (55.6 - x) (0.46) = (10.0) (x - 18.0) (0.54)

9.2 (55.6 - x) = 5.4 (x - 18)

511.52 - 9.2x = 5.4x - 97.2

14.6x = 608.72

x = 41.7 °C


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Worksheet #2